Wordle Character Compare

Given a guess and a target word of equal length, color each guess letter: green for correct letter and position, yellow for correct letter wrong position, gray otherwise — but obey Wordle's nuanced "each target letter can be matched at most once" rule. Requirements Given two equal-length lowercase strings guess and target, return an array of colors (one per character of guess): GREEN if guess[i] == target[i]. YELLOW if guess[i] != target[i] but guess[i] exists somewhere in target and has not yet been consumed by a green or earlier yellow match. GRAY otherwise. Function signature: Color[] wordle(String guess, String target) The trap interviewers explicitly probe: when guess has multiple copies of a letter and target has fewer, only the first ones up to the target count are colored (green takes priority, then yellow left-to-right). The rest are gray. The naive "set-based" check fails this case. Examples guess = 'lemon' target = 'lover' L: green E: yellow M: gray O: yellow N: gray guess = 'speed' target = 'erase' S: yellow P: gray E: yellow (no position matches, so no green; target has two e's) E: yellow (second e still available in target) D: gray guess = 'array' target = 'radar' A: yellow (a exists, not at this position) R: yellow (target 'radar' has two r's) R: yellow (second r still available) A: green (guess 'a' == target 'a' at this index) Y: gray Notes Two-pass algorithm: first pass marks greens and decrements a Counter(target) for each green letter. Second pass scans the remaining un-colored positions and assigns yellow if the counter is still positive, gray otherwise; decrement on yellow. One-pass is possible but the two-pass version is clearer for interview narration and easier to verify by hand. Time and space O(n) where n is the word length (alphabet size σ is constant). The most common bug is doing only the yellow pass first or using a set instead of a counter — both produce wrong colors on duplicate-letter inputs. A follow-up sometimes asks for the rule set as an enum and a struct-of-arrays output. Mostly cosmetic; mention before coding. Preparation Hand-trace the three examples above (especially the duplicate-letter cases) before writing code — this is the single largest source of bugs. Implement the two-pass algorithm and one one-pass alternative; argue why the two-pass version is more defensible. Drill the rule clarification out loud: confirm green priority, then left-to-right yellow consumption, then everything else is gray.